Is it true that if the order of numerator of a system function is larger than its denominator, then we can say that the system is causal? (Without finding R.O.C.)
Is there similar thing for Stability?
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$\begingroup$ Have you looked at Why more poles than zeroes? It looks duplicate to me. $\endgroup$– Valentin TihomirovCommented Mar 15, 2016 at 7:05
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2 Answers
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When $H(z)$ is rational, then the system is causal if and only if its ROC is the exterior of a circle outside the out-most pole, and the order of numerator is no greater than the order of the denominator.
A causal LTI system with a rational transfer function $H(z)$ is stable if and only if all poles of $H(z)$ are inside the unit circle of the z-plane, i.e., the magnitudes of all poles are smaller than 1.
so you also need to find the R.O.C.
answered Mar 15, 2016 at 6:32
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for causal systems (which are the only ones i know that are physically realizable) which means the number of zeros may not exceed the number of poles, the system is stable if and only if all poles are inside of the unit circle. that's the only criteria you need to worry about.
there is a discrete-time counterpart to the Routh-Hurwitz criterion called the Jury test that allows you to determine stability without factoring the denominator of $H(z)$ into factors that identify the poles, but the basic criterion is the same. whether you factor the denominator into factors with poles $(z-p_i)$ or use the Jury test, the salient test is "are there any poles, $p_i$, such that $|p_i| \ge 1$?" if so, the system is not stable. if all poles satisfy $|p_i| < 1$, the system is stable (we must include any poles that are cancelled by zeros).
answered Mar 15, 2016 at 18:32
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$\begingroup$ I think it's clearer to state the sentence "the number of zeros may not exceed the number of poles" as "the degree of the numerator polynomial may not exceed the degree of the denominator polynomial", because the first statement is wrong if we include poles at infinity (and why shouldn't we?). E.g., $$H(s)=\frac{(s+1)(s+2)}{s+3}$$ has two zeros, but also two poles (one at $s=-3$ and one at $s=\infty$). So the number of zeros doesn't exceed the number of poles, yet the system is not causal and stable. $\endgroup$– Matt L.Commented Mar 15, 2016 at 20:17
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$\begingroup$ @MattL, you're right about the specific fact, but your example is problematic. how are you going to implement your $H(s)$ with integrators? if you try to multiply both numerator and denominator with $s^{-2}$ and then represent the missing factor (and pole at $\infty$) with $(1-p z^{-1})$, you will have a scaling problem with the whole transfer function. so, in fact, if the number of zeros exceeds the number of poles, i see realization as problematic. perhaps divide out one $s$ and implement that as a differentiator summed to the output of a 1 pole, 1 zero system. $\endgroup$ Commented Mar 17, 2016 at 20:15
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$\begingroup$ Nobody can implement that system because it's neither causal nor stable. But the point is that its number of poles equals its number of zeros (if poles at infinity are counted). This was just to show that you can have a non-causal system even if the number of zeros does not exceed the number of poles. So it's easier and clearer to talk about the degrees of numerator and denominator instead of the number of poles and zeros. $\endgroup$– Matt L.Commented Mar 17, 2016 at 20:51
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$\begingroup$ how many poles at $\infty$ are you gonna toss in there? why not 2 or 4 or some other contrived number? $\endgroup$ Commented Mar 18, 2016 at 0:37
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$\begingroup$ No, no need for that. The wickedly contrived number '1' was sufficient to prove my point. $\endgroup$– Matt L.Commented Mar 18, 2016 at 8:10